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solve-x-x-3-5-




Question Number 109193 by mr W last updated on 21/Aug/20
solve   x^x^3  =5
solvexx3=5
Commented by aurpeyz last updated on 21/Aug/20
 x^x^3  =2^2   x^3 =2  x=^3 _ (√2)
xx3=22x3=2x=32
Commented by aurpeyz last updated on 21/Aug/20
i just tried. pls correct me
ijusttried.plscorrectme
Commented by mr W last updated on 21/Aug/20
if x^3 =2 then x=2, impossble.
ifx3=2thenx=2,impossble.
Commented by mr W last updated on 21/Aug/20
i have changed the question to  x^x^3  =5
ihavechangedthequestiontoxx3=5
Commented by aurpeyz last updated on 21/Aug/20
i wrote cube root of 2
iwrotecuberootof2
Commented by aurpeyz last updated on 21/Aug/20
okay sir. i will try
okaysir.iwilltry
Commented by mr W last updated on 21/Aug/20
you wrote  x^x^3  =2^2   if x^3 =2, then you mean also x=2.
youwrotexx3=22ifx3=2,thenyoumeanalsox=2.
Commented by aurpeyz last updated on 21/Aug/20
yes. i am wrong. if the indeces are equal. the base  will be equal.
yes.iamwrong.iftheindecesareequal.thebasewillbeequal.
Answered by Dwaipayan Shikari last updated on 21/Aug/20
x^3 logx=log5  3logx e^(3logx) =3log5  3logx=W_0 (3log5)  x=e^((W_0 (3log5))/3) =1.5459..
x3logx=log53logxe3logx=3log53logx=W0(3log5)x=eW0(3log5)3=1.5459..
Commented by mr W last updated on 21/Aug/20
good!
good!
Answered by Aziztisffola last updated on 21/Aug/20
ln(x^x^3  )=ln(5) ⇔x^3 ln(x)=ln(5)  ⇔ln(x)e^(3ln(x)) =ln(5)  ⇔3ln(x)e^(3ln(x)) =3ln(5)   W(3ln(x)e^(3ln(x)) )=W(3ln(5))   3ln(x)=W(3ln(5))   x=e^((W(3ln(5)))/3)
ln(xx3)=ln(5)x3ln(x)=ln(5)ln(x)e3ln(x)=ln(5)3ln(x)e3ln(x)=3ln(5)W(3ln(x)e3ln(x))=W(3ln(5))3ln(x)=W(3ln(5))x=eW(3ln(5))3
Commented by aurpeyz last updated on 24/Aug/20
Pls what is the w or w_0  being inserted?
Plswhatistheworw0beinginserted?
Commented by Aziztisffola last updated on 24/Aug/20
The Lambert W function; the invers    function of  f:x ∣→xe^x    W=f^( −1 )
TheLambertWfunction;theinversfunctionoff:xxexW=f1
Commented by aurpeyz last updated on 25/Aug/20
wow. that was why you multiplied through  by 3?
wow.thatwaswhyyoumultipliedthroughby3?
Answered by 1549442205PVT last updated on 22/Aug/20
Put x^3 =y⇒x=^3 (√y)⇒^3 ((√y))^y =5  yln(^3 (√y))=ln5⇔(y/3)lny=ln5  ylny=3ln5⇔y=e^((3ln5)/y) ⇔(1/y).e^((3ln5)/y) =1  Put (1/y)=z we get ze^(z.3ln5) =1  ⇔z.3ln5ln5e^(z.3ln5) =3ln5  ⇒z.3ln5=W_0 (3ln5)=1.306568..  ⇒z=1.306568/(3ln5)=0.27060.  ⇒y=1/z=3.695417  ⇒x=^3 (√y)=1.5460
Putx3=yx=3y3(y)y=5yln(3y)=ln5y3lny=ln5ylny=3ln5y=e3ln5y1y.e3ln5y=1Put1y=zwegetzez.3ln5=1z.3ln5ln5ez.3ln5=3ln5z.3ln5=W0(3ln5)=1.306568..z=1.306568/(3ln5)=0.27060.y=1/z=3.695417x=3y=1.5460

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